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Using Circuit Wizard 3 I tested the forward bias of a 1N4001 diode. This is the circuit I used:

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and this is the graph I got when I plotted current through the ammeter against voltage across the diode.

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Why is it that the current voltage graph gives an exponential curve?

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  • \$\begingroup\$ Why is it that the current voltage graph gives an exponential curve? What curve did you expect and why ? Google for "diode curve" what do you see ? \$\endgroup\$ – Bimpelrekkie Oct 17 '16 at 9:37
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    \$\begingroup\$ It's commendable that you did the experiment. They are modeled, forward biased, as \$I=I_S\cdot\left(e^{\frac{V q}{n k T}}-1\right)\$. Were you looking for just a statement that it's okay that you found such a curve? Or did you want a discussion at a solid state physics level? \$\endgroup\$ – jonk Oct 17 '16 at 9:43
  • \$\begingroup\$ If the curve was linear, you'd call it resistor, not diode. \$\endgroup\$ – Dmitry Grigoryev Oct 17 '16 at 9:43
  • \$\begingroup\$ I expected an exponential curve given that I looked at current voltage graphs for diodes before running the simulation. My question is why the curve would be exponential. I understand that in forward bias the holes and electrons combine allowing current to flow. \$\endgroup\$ – Darth Vader Oct 17 '16 at 9:44
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    \$\begingroup\$ I recommend Jacob Millman's "Microelectronics," and just read the first few chapters. It's easy to follow. (I still have my 1979 edition here, which is quite good. Should be cheap to buy an older copy. There are new editions, but they are co-authored, cover more, and actually might be more of a stretch for you. So I'd recommend an earlier edition, for now.) \$\endgroup\$ – jonk Oct 17 '16 at 9:47

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